What is the integral of e^{-4x}sin(5x) ?

The integral of e^{-4x}sin(5x) is -(5e^{-4x}cos(5x))/(41)-(4e^{-4x}sin(5x))/(41)+C

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integral of e^{-4x}sin(5x)

\int e^{- 4 x} sin (5 x) d x

- \frac{5 e ^{- 4 x} cos ( 5 x )}{41} - \frac{4 e ^{- 4 x} sin ( 5 x )}{41} + C

Solution steps

\int e^{- 4 x} sin (5 x) d x

Apply Integration By Parts

: - \frac{1}{5} e^{- 4 x} cos (5 x) - \int \frac{4}{5} e^{- 4 x} cos (5 x) d x

= - \frac{1}{5} e^{- 4 x} cos (5 x) - \int \frac{4}{5} e^{- 4 x} cos (5 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{5} e^{- 4 x} cos (5 x) - \frac{4}{5} \cdot \int e^{- 4 x} cos (5 x) d x

Apply Integration By Parts

: \frac{1}{5} e^{- 4 x} sin (5 x) - \int - \frac{4}{5} e^{- 4 x} sin (5 x) d x

= - \frac{1}{5} e^{- 4 x} cos (5 x) - \frac{4}{5} (\frac{1}{5} e^{- 4 x} sin (5 x) - \int - \frac{4}{5} e^{- 4 x} sin (5 x) d x)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{5} e^{- 4 x} cos (5 x) - \frac{4}{5} (\frac{1}{5} e^{- 4 x} sin (5 x) - (- \frac{4}{5} \cdot \int e^{- 4 x} sin (5 x) d x))

Therefore

\int e^{- 4 x} sin (5 x) d x = - \frac{1}{5} e^{- 4 x} cos (5 x) - \frac{4}{5} (\frac{1}{5} e^{- 4 x} sin (5 x) - (- \frac{4}{5} \cdot \int e^{- 4 x} sin (5 x) d x))

Isolate

\int e^{- 4 x} sin (5 x) d x

= - \frac{5 e ^{- 4 x} cos ( 5 x )}{41} - \frac{4 e ^{- 4 x} sin ( 5 x )}{41}

Add a constant to the solution

= - \frac{5 e ^{- 4 x} cos ( 5 x )}{41} - \frac{4 e ^{- 4 x} sin ( 5 x )}{41} + C

f (x) = \frac{110}{x ^{12}}

f (x) = 7 x^{4} arctan (4 x^{4})

\int \frac{7}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (2 d x)

x \to 1 lim (4 x^{3} - x^{2} + 7 x)

What is the integral of e^{-4x}sin(5x) ?
The integral of e^{-4x}sin(5x) is -(5e^{-4x}cos(5x))/(41)-(4e^{-4x}sin(5x))/(41)+C